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Bingel [31]
3 years ago
5

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.40 moles moles of magnesium perchlorate, Mg(ClO4)2Mg

(ClO4)2. Express the number of moles of MgM, Cl Cl, and OO atoms numerically, separated by commas.
Chemistry
1 answer:
nexus9112 [7]3 years ago
4 0

Answer:Moles of Mg, Cl,  0 =  5.40, 10.80, 43.2

Explanation:

1  mole of Mg(ClO4)2  contains  1 mole of Magnesium, Mg atoms

                                                    2 Moles of Chlorine, Cl atoms

                                                    8 Molresof Oxygen atoms

Such that to find the number of moles of each element, We have that

------1 mole Magnesum  atoms/ 1 mol of  Mg(ClO4)2   x 5.40 moles  of magnesium perchlorate, Mg(ClO4)2 = 1 x 5.40 =5.40 moles

------2 moles ofChlorine atoms/ 1 mol of Mg(ClO4)2   x 5.40 moles of magnesium perchlorate, Mg(ClO4)2 = 2 x 5.40 = 10.80 moles

------8 moles Oxygen atoms/ 1 mol of  Mg(ClO4)2   x 5.40 moles  of magnesium perchlorate, Mg(ClO4)2 = 8 x 5.40 = 43.2 moles

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If I added 50 mL of a 1.0 M HCIO2 solution to 50 mL of a 1.0 M NaOH
kozerog [31]

Answer:

pH=7

Explanation:

0.05 L*1.0M= 0.05 mol HClO2  and 0.05 mol NaOH      

HClO2 +     NaOH ---> NaClO2 + H2O

0.05 mol     0.05 mol

As we can see acid and base  react completely,  solution will be neutral, so pH =7

3 0
3 years ago
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydr
Semmy [17]

Answer:

K⁺ (aq)  +  F⁻ (aq)  +  H⁺ (aq)  +  Cl⁻ (aq)  → KCl (aq) + H⁺ (aq)  +  F⁻ (aq)

Explanation:

KF (aq) +  HCl (aq) →  KCl (aq)  + HF (aq)

KF (aq) → K⁺ (aq) +  F⁻ (aq)

HCl (aq) →  H⁺ (aq)  +  Cl⁻ (aq)

KCl (aq) → K⁺ (aq) +  Cl⁻ (aq)

HF (aq) →  H⁺ (aq)  +  F⁻ (aq)

7 0
3 years ago
Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G
seraphim [82]

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g     (Conversion factor:  1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = \frac{100}{39}\times 0.1=0.256g of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

8 0
3 years ago
Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K.
Elis [28]

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

\Delta G= \Delta H - T * \Delta S

<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>

Calculating the \Delta G= -1267 - 473 K* \Delta S :

\Delta G= -1267 - 473 K* \Delta S

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: \Delta S>0[/tex}Back to this expression: [tex]\Delta G= -1267 - 473 K* \Delta S

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

4 0
3 years ago
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
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